3 ms·

Given that the blocks are connected only by gravity, how could that bottom block possibly help?

• m4dc4pXXX 10y ago
  My thought too.
• ghusbands 10y ago
  If you're talking about the one in the first diagram, it indeed doesn't help. One could argue that the diagram is just before the tower is shifted, as described in the iterative algorithm.
• LeifCarrotson 10y ago
  It's only used to for the logical induction. It doesn't help the blocks above balance any better. The bottom block hanging over the table with a heavy weight on the top right edge means the whole system is prone to tipping, and moving the bottom block back so we're just worried about the next edge higher up means that the system is, on average, less prone to tip over. It turns out to be easy to calculate the difference in tippyness/torque between these two cases, so we can use this to calculate how far the block can hang over. One block obviously balances at 1/2, and has a center of mass at the edge. Two blocks obviously still have a center of mass at the edge - else they would tip, or could go closer to the edge without tipping. But how far out is this second block hanging? If we put the second block flush with the edge of the table, it has a center of mass at -1/2 and a mass of 1, and the top block has a center of mass at 0 and a mass of 1. So the center of mass of the system is at -1/4 and we can shift it this far to the right. With 3 blocks, we have a mass of 2 blocks with the center at position 0, and another block below with a center of mass at -1/2 and a mass of 1, for a system centered at -1/6th (and so on for more blocks). We always shift it over in the end, but it makes the math of how far to shift it easier to compute if we start with it backed up a bit.